Presented at the COMSOL Conference 2010 India
Two Dimensional FEM
Simulation of Ultrasonic Wave
Propagation in Isotropic Solid
Media Using COMSOL®
Bikash Ghose1 *, Krishnan Balasubramaniam2
C V Krishnamurthy3, A Subhananda Rao1
#
1
High Energy Materials Research Laboratory, Sutarwadi, Pune - 21
2 Center for Non Destructive Evaluation, IIT Madras, Chennai - 36
3 Department of Physics, IIT Madras, Chennai -36
E-mail: * [email protected] , # [email protected]
Motivation
Application of Ultrasonic wave propagation
•
•
•
•
•
Under Water Acoustics
Medical Diagnostic
Structural Health Monitoring (SHM)
Non-Destructive Evaluation (NDE)
Material Characterization
Non-Destructive Evaluation
• Bulk wave
Longitudinal wave (Pulse echo technique, Through
transmission etc)
Transverse Wave
• Surface Wave
• Guided Wave / Lamb Wave etc.
Guided Wave
• Many different modes of ultrasonic vibration
(Symmetric, Anti symmetric)
Numerical Simulation of Wave propagation
• Fluid (Only Longitudinal wave)
• Solid (Longitudinal, Transverse, Surface, Lamb
etc)
Use of COMSOL for simulation of wave
propagation in solids
Transient Analysis for Wave
Propagation
Transient Analysis always pose difficult problems
Length of element (∆x)
Time steps (∆t)
Type of Elements (Triangular, Quad)
Integration scheme for integration over time
Issues
•
•
•
Instability
Numerical Dispersion
Convergence
Numerical Simulation of Ultrasonic
Wave Propagation using COMSOL®
Thumb Rule ?
•Length of
Element (∆x) = λ/12
•Time Step = ∆x/Cph
Initial Displacement
Solution at 7.0×10-5 s
Displacement at diff positions with time
Solution at 1.7×10-4 s
Materials & Ultrasonic Wave
Properties
x 10
6
4
D isplac em ent (m )
Young’s Modulus (E) = 2 ×
Pa
Poisson’s ratio (ν) = 0.33
Density (ρ) = 7850 kg/m3
Ultrasonic velocity (CL) (longitudinal
wave) = 5850 m/s
Frequency of incident ultrasonic wave (f)
= 20 kHz = 20×103 s-1
Wavelength of the longitudinal ultrasonic
wave (λL) = CL/f = 0.2925 m
Source Length : 0.04 m (40 mm)
(located at the middle)
Source Excitation : On a line and points
on line
Simulation Domain : 5.0 m (L) x 2.5 m
(H)
8
2
0
-2
-4
-6
-8
-10
0
0.5
1
1.5
-4
x 10
Time (s)
x 10
Single-Sided Amplitude Spectrum of y(t)(Hanning windowed signal)
-7
5
4.5
4
3.5
3
|Y(f)|
-7
1011
2.5
2
1.5
1
0.5
0
2
4
6
8
10
Frequency (Hz)
12
14
16
18
20
x 10
4
Meshing and Application Modes
Triangular Elements
Elements Automatically
Generated
Approximation : Plane Stain
Element Type : Lagrange
Quadratic
Analysis : Transient
Solver : Time Dependent
Effect of Length of Element (∆x)
As per CFL Criteria
∆tcritical =
∆xmax ∆xmax 0.0182m
=
=
= 3.1×10−6 s
C ph
CL
5850m / s
The time steps chosen initially for simulation for different ∆xmax is
2.5×10-6 s
Onward Propagating Wave at various
Distances for Different λL/∆xmax at t = 4×10-4 s
Ratio = 2
Ratio = 3
Ratio = 4
Ratio= 5
Ratio = 8
Ratio = 9
Ratio = 12
Ratio = 16
At 0.3 m
8.00E-08
Displacements (m)
6.00E-08
4.00E-08
2.00E-08
0.00E+00
-2.00E-08
0
0.0001
0.0002
0.0003
0.0004
0.0005
At 0.5 m
6.00E-08
Ratio = 2
Ratio = 3
Ratio = 4
Ratio = 5
Ratio = 8
Ratio = 9
Ratio = 12
Ratio = 16
4.00E-08
Displacement (m)
1.00E-07
2.00E-08
0.00E+00
0
0.0001
0.0002
0.0003
0.0004
0.0005
-2.00E-08
-4.00E-08
-4.00E-08
-6.00E-08
-8.00E-08
-6.00E-08
Time (s)
Time (t)
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
4.00E-08
3.00E-08
Displacement (m)
2.00E-08
1.00E-08
=2
=3
=4
=5
=8
=9
= 12
= 16
0.00E+00
0
0.0001
0.0002
0.0003
-1.00E-08
0.0004
0.0005
At 1.0 m
4.00E-08
3.00E-08
2.00E-08
Displacement
At 0.8 m
5.00E-08
1.00E-08
0.00E+00
0
0.00005 0.0001 0.00015 0.0002 0.00025 0.0003 0.00035 0.0004 0.00045
-1.00E-08
-2.00E-08
-2.00E-08
-3.00E-08
-3.00E-08
-4.00E-08
-4.00E-08
-5.00E-08
Time (s)
Ratio = 2
Ratio = 3
Ratio = 4
Ratio = 5
Ratio = 8
Ratio = 9
Ratio = 12
Ratio = 16
Time (t)
• Oscillations remains after passing by of signal
• Convergence of the solution for the ratio (λL /∆xmax) ≥ 8
• Time steps are taken from solver or exactly what has been
given does not make any difference to the propagating signal
Onward Propagating Wave for Different
λL/∆xmax at t = 4×10-4 s
Ratio = 2
Ratio = 3
ratio = 4
ratio = 5
Ratio = 8
Ratio = 9
Ratio = 12
2.50E-08
2.00E-08
Displacement (m)
1.50E-08
1.00E-08
5.00E-09
0.00E+00
-5.00E-09
0
0.5
1
1.5
2
2.5
3
-1.00E-08
-1.50E-08
-2.00E-08
-2.50E-08
Distance from Source center (m)
Line profile (Displacement Vs Distance from source) at
0.0004s for different wavelength to element length ratio
Effect of Time Steps (∆t)
As per CFL criteria the time steps should be less than ∆x/CL
for example, for the maximum element length of λL /10 the
time steps should be ≤ (λL /10×CL) that means if
∆xmax =
λL
10
∆x
λL
⇒ ∆t ≤ max =
CL
10 × C L
⇒ ∆t ≤
1
10 × f
Signal so far obtained is not as expected
Major change in the shape of signal
Time steps further decreased and solution was checked for
convergence
Case: I
∆t = 10 ×10−6 s
∆t = 1.0 × 10 −6 s
λL/∆xmax= 5
∆t = 5 ×10−6 s
∆t = 2.0 × 10 −6 s
∆t = 0.5 × 10 −6 s
∆t = 0.2 ×10−6 s
Case: II
∆t = 6.26 ×10−6 s
∆t = 1.0 × 10 −6 s
λL/∆xmax= 8
∆t = 5 ×10−6 s
∆t = 0.5 × 10 −6 s
∆t = 2.0 × 10 −6 s
∆t = 0.2 ×10−6 s
Case: III
∆t = 5 ×10−6 s
∆t = 1.0 × 10 −6 s
λL/∆xmax= 12
∆t = 2.5 × 10−6 s
∆t = 0.5 × 10 −6 s
Effect of Excitation on Line and Points on
Line for λL/∆xmax= 8 and ∆t = 2.5x10-6 s
(Line source)
At 1.0 m
At 0.5 m
(Points on Line source)
At 1.0 m
At 0.5 m
Effect of Excitation on Line and Points on
Line for λL/∆xmax= 8 and ∆t = 2.5x10-6 s
Line profile at t = 4x10-4 s
(Line source)
(Points on Line source)
Time and Frequency Domain Signal for
Forward Propagating Wave
λL
∆xmax
λL
=8
∆xmax
8.00E-08
4.00E-08
4.00E-08
4.00E-08
0.00E+00
0.00E+00
2.00E-04
4.00E-04
6.00E-04
-4.00E-08
0.00E+00
0.00E+00
2.00E-04
-4.00E-08
2.00E-04
14
6.00E-04
-1.20E-07
Time (s)
Single-Sided Amplitude Spectrum
-9
Single-Sided Amplitude Spectrum
4.00E-04
-4.00E-08
Time (s)
Time (s)
-9
0.00E+00
0.00E+00
=5
-8.00E-08
-1.20E-07
-1.20E-07
x 10
6.00E-04
-8.00E-08
-8.00E-08
10
4.00E-04
Displacem ent (m)
8.00E-08
Displacement (m)
8.00E-08
x 10
Single-Sided Amplitude Spectrum
-9
x 10
9
12
12
10
10
8
8
7
|Y(f)|
6
5
|Y (f)|
8
|Y (f)|
D isplacem ent (m )
∆xmax
λL
= 12
4
6
3
4
4
2
2
6
2
1
0
1
2
3
4
5
Frequency (Hz)
6
7
8
9
1
x 10
2
4
All signals are for ∆t =0.5x10-6 s
3
4
5
6
Frequency (Hz)
7
8
9
4
x 10
1
2
3
4
5
6
Frequency (Hz)
7
8
9
x 10
4
Time and Frequency Domain Signal for
Back Wall Reflected Wave
λL
∆xmax
∆xmax
4.00E-08
2.00E-08
2.00E-08
2.00E-08
0.00E+00
0.00E+00
4.00E-04
8.00E-04
1.20E-03
-2.00E-08
0.00E+00
0.00E+00
4.00E-04
-4.00E-08
-6.00E-08
-6.00E-08
4.00E-04
x 10
6
5
|Y(f)|
4
8.00E-04
1.20E-03
-2.00E-08
-6.00E-08
Time (s)
Time (s)
Single-Sided Amplitude Spectrum
-9
0.00E+00
0.00E+00
=5
-4.00E-08
Single-Sided Amplitude Spectrum
-9
Single-Sided Amplitude Spectrum
-9
x 10
8
8
7
7
6
6
5
5
|Y(f)|
x 10
1.20E-03
-2.00E-08
-4.00E-08
7
8.00E-04
D isplacem ent (m )
4.00E-08
Time (s)
|Y(f)|
λL
=8
4.00E-08
Displacement (m)
Displacement (m)
∆xmax
λL
= 12
4
4
3
2
1
0
1
2
3
4
5
6
Frequency (Hz)
7
8
9
3
3
2
2
1
1
1
2
4
x 10
All signals are for ∆t =0.5x10-6 s
3
4
5
6
Frequency (Hz)
7
8
1
9
x 10
4
2
3
4
5
6
Frequency (Hz)
7
8
9
4
x 10
Simulation Results for λL/∆xmax= 8
∆t = 0.5 x 10-6 s at different time
At t = 1x10-4 s
Simulation Results for λL/∆xmax= 8
∆t = 0.5 x 10-6 s at different time
At t = 4x10-4 s
Simulation Results for λL/∆xmax= 8
∆t = 0.5 x 10-6 s at different time
At t = 5x10-4 s
Simulation Results for λL/∆xmax= 8
∆t = 0.5 x 10-6 s at different time
At t = 8x10-4 s
Conclusions
Wave Propagation can well be modeled using
COMSOL with excitation on line segment and / or
on points on line (Difference only in the
maximum displacements)
λL/∆xmax ≥ 8 irrespective of any time steps less
than calculated by CFL criteria ( For triangular
free meshing)
Time step ∆t should be about T/100 even if λL/∆x
≈ 16
No substantial difference in frequency content of
forward as well as back wall reflected ultrasonic
signal observed
E-mail: [email protected] , [email protected]